### 3.44 $$\int e^{-2 \coth ^{-1}(a x)} x \, dx$$

Optimal. Leaf size=25 $\frac{2 \log (a x+1)}{a^2}-\frac{2 x}{a}+\frac{x^2}{2}$

[Out]

(-2*x)/a + x^2/2 + (2*Log[1 + a*x])/a^2

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Rubi [A]  time = 0.0323883, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.3, Rules used = {6167, 6126, 77} $\frac{2 \log (a x+1)}{a^2}-\frac{2 x}{a}+\frac{x^2}{2}$

Antiderivative was successfully veriﬁed.

[In]

Int[x/E^(2*ArcCoth[a*x]),x]

[Out]

(-2*x)/a + x^2/2 + (2*Log[1 + a*x])/a^2

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6126

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int e^{-2 \coth ^{-1}(a x)} x \, dx &=-\int e^{-2 \tanh ^{-1}(a x)} x \, dx\\ &=-\int \frac{x (1-a x)}{1+a x} \, dx\\ &=-\int \left (\frac{2}{a}-x-\frac{2}{a (1+a x)}\right ) \, dx\\ &=-\frac{2 x}{a}+\frac{x^2}{2}+\frac{2 \log (1+a x)}{a^2}\\ \end{align*}

Mathematica [A]  time = 0.012387, size = 25, normalized size = 1. $\frac{2 \log (a x+1)}{a^2}-\frac{2 x}{a}+\frac{x^2}{2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x/E^(2*ArcCoth[a*x]),x]

[Out]

(-2*x)/a + x^2/2 + (2*Log[1 + a*x])/a^2

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Maple [A]  time = 0.039, size = 24, normalized size = 1. \begin{align*} -2\,{\frac{x}{a}}+{\frac{{x}^{2}}{2}}+2\,{\frac{\ln \left ( ax+1 \right ) }{{a}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x+1)*(a*x-1),x)

[Out]

-2*x/a+1/2*x^2+2*ln(a*x+1)/a^2

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Maxima [A]  time = 1.05236, size = 35, normalized size = 1.4 \begin{align*} \frac{a x^{2} - 4 \, x}{2 \, a} + \frac{2 \, \log \left (a x + 1\right )}{a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a*x-1)/(a*x+1),x, algorithm="maxima")

[Out]

1/2*(a*x^2 - 4*x)/a + 2*log(a*x + 1)/a^2

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Fricas [A]  time = 1.84129, size = 59, normalized size = 2.36 \begin{align*} \frac{a^{2} x^{2} - 4 \, a x + 4 \, \log \left (a x + 1\right )}{2 \, a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a*x-1)/(a*x+1),x, algorithm="fricas")

[Out]

1/2*(a^2*x^2 - 4*a*x + 4*log(a*x + 1))/a^2

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Sympy [A]  time = 0.260727, size = 20, normalized size = 0.8 \begin{align*} \frac{x^{2}}{2} - \frac{2 x}{a} + \frac{2 \log{\left (a x + 1 \right )}}{a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a*x-1)/(a*x+1),x)

[Out]

x**2/2 - 2*x/a + 2*log(a*x + 1)/a**2

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Giac [A]  time = 1.11769, size = 41, normalized size = 1.64 \begin{align*} \frac{a^{2} x^{2} - 4 \, a x}{2 \, a^{2}} + \frac{2 \, \log \left ({\left | a x + 1 \right |}\right )}{a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a*x-1)/(a*x+1),x, algorithm="giac")

[Out]

1/2*(a^2*x^2 - 4*a*x)/a^2 + 2*log(abs(a*x + 1))/a^2